1195: For Group with Topology with Continuous Operations (Especially, Topological Group), Neighborhood of 1, and Positive Natural Number, There Is Symmetric Neighborhood of 1 Whose Power to Natural Number Is Contained in Neighborhood

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description/proof of that for group with topology with continuous operations (especially, topological group), neighborhood of $1$, and positive natural number, there is symmetric neighborhood of $1$ whose power to natural number is contained in neighborhood

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Topics

About: group
About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of group.
• The reader knows a definition of topological space.
• The reader knows a definition of continuous map.
• The reader knows a definition of neighborhood of point on topological space.
• The reader knows a definition of symmetric subset of group.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.
• The reader admits the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

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Target Context

• The reader will have a description and a proof of the proposition that for any group with any topology with any continuous operations (especially, topological group), any neighborhood of $1$, and any positive natural number, there is a symmetric neighborhood of $1$ whose power to the natural number is contained in the neighborhood.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$G$: $\in \{\text{ the groups }\}$ with any topology such that the group operations are continuous
$N_1^{\prime }$: $\in \{\text{ the neighborhoods of }1\text{ on }G\}$
$n$: $\in \mathbb{N}\setminus \{0\}$
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Statements:
$\mathrm{\exists }{N}_1\in \{\text{ the symmetric neighborhoods of }1\text{ on }G\}(N_1^n\subseteq N_1^{\prime })$
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2: Note

We do not have any immediate reason why $G$ should not be Hausdorff (so, $G$ should not be any topological group), but we use this proposition for proving that a $G$ is Hausdorff, so, Hausdorff-ness is not presupposed.

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3: Proof

Whole Strategy: Step 1: think of the continuous $n$-multiplication map, $f_n:G\times ...\times G\to G$, and take an open neighborhood of $1$, $U_1$, such that $f_n(U_1\times ...\times U_1)\subseteq N_1^{\prime }$; Step 2: take a symmetric neighborhood of $1$, $N_1$, such that $N_1\subseteq U_1$, and see that $N_1^n=f(N_1\times ...\times N_1)\subseteq N_1^{\prime }$.

Step 1:

Let us think of the $n$-multiplication map, $f_n:G\times ...\times G\to G,(g_1,...,g_n)\mapsto g_1...g_n$.

$f_n$ is continuous, by the proposition that for any group with any topology with any continuous operations (especially, topological group), any finite multiplication map is continuous.

So, as $f_n(1,...,1)=1$, there is an open neighborhood of $(1,...,1)\in G\times ...\times G$, $U_{(1,...,1)}\subseteq G\times ...\times G$, such that $f_n(U_{(1,...,1)})\subseteq N_1^{\prime }$.

There is an open neighborhood of $1$, $U_1\subseteq G$, such that $(1,...,1)\in U_1\times ...\times U_1\subseteq U_{(1,...,1)}$, by the definition of product topology: while there are some open neighborhoods of $1$, $U_{1,1},...U_{1,n}\subseteq G$, such that $(1,...,1)\in U_{1,1}\times ...\times U_{1,n}\subseteq U_{(1,...,1)}$, we can take $U_1:=U_{1,1}\cap ...\cap U_{1,n}$.

Step 2:

There is a symmetric neighborhood of $1$, $N_1$, such that $N_1\subseteq U_1$, by the proposition that for any group with any topology with any continuous operations (especially, topological group), the set of the symmetric neighborhoods of $1$ is a neighborhood basis at $1$.

$f_n(N_1\times ...\times N_1)\subseteq f_n(U_1\times ...\times U_1)\subseteq f_n(U_{(1,...,1)})\subseteq N_1^{\prime }$.

Let us see that $f_n(N_1\times ...N_1)=N_1^n$.

For each $g\in f_n(N_1\times ...\times N_1)$, $g=f_n((g_1,...,g_n))=g_1...g_n\in N_1^n$; for each $g\in N_1^n$, $g=g_1...g_n=f_n((g_1,...,g_n))\in f_n(N_1\times ...\times N_1)$.

So, $N_1^n=f(N_1\times ...\times N_1)\subseteq N_1^{\prime }$.

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References

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